Question

ajutor !!!! ofer 30 pct!!!

Answer 1

$a) \frac{5 - x}{2} - \frac{1 - 3x}{5} = 4 \: | \times 10$

$5(5 - x) - 2(1 - 3x) = 40$

$25 - 5x - 2 + 6x = 40$

$- 5x + 6x = 40 - 25 + 2$

$x = 17 \: \in \: \mathbb{R}$

$b) \frac{x - 1}{2} - \frac{2x - 1}{4} + 2 = x + 3 \: | \times 4$

$2(x - 1) - 2x + 1 + 8 = 4x + 12$

$2x - 2 - 2x + 1 + 8 = 4x + 12$

$2x - 2x - 4x = 12 + 2 - 1 - 8$

$- 4x = 5 \: | \times ( - 1)$

$4x = - 5$

$x = - \frac{5}{4} \: \in \: \mathbb{R}$

$c)2(x - 1) - 3(x + 2) = 5$

$2x - 2 - 3x - 6 = 5$

$2x - 3x = 5 + 2 + 6$

$- x = 13 \: | \times ( - 1)$

$x = - 13 \: \in \: \mathbb{R}$

$d)3(x + 1) - 4(x - 2) = 6$

$3x + 3 - 4x + 8 = 6$

$3x - 4x = 6 - 3 - 8$

$- x = - 5 \: | \times ( - 1)$

$x = 5 \: \in \: \mathbb{R}$

$e)3(x - 2) - ( - 6 + x) \times ( - 2) = - 18$

$3x - 6 - (12- 2x) = - 18$

$3x - 6 -12+ 2x = - 18$

$3x + 2x = - 18+6+12$

$5x = 0$

$x = 0\: \in \: \mathbb{R}$

$f) {(x - 1)}^{2} - (x + 2)( x - 1) + 1 = 0$

${x}^{2} - 2x + 1 - ( {x}^{2} - x + 2x - 2) + 1 = 0$

${x}^{2} - 2x + 1 - ( {x}^{2} + x - 2) + 1 = 0$

${x}^{2} - 2x + 1 - {x}^{2} - x + 2 + 1 = 0$

${x}^{2} - {x}^{2} - 2x - x = - 1 - 2 - 1$

$- 3x = - 4 \: | \times ( - 1)$

$3x = 4$

$x = \frac{4}{3} \: \in \: \mathbb{R}$